Homework_10 - Nguyen, Thanh Homework 10 Due: Oct 9 2007,...

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Unformatted text preview: Nguyen, Thanh Homework 10 Due: Oct 9 2007, 7:00 pm Inst: D Weathers 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A 2 . 83 g bullet moving at 270 m / s penetrates a tree to a depth of 6 . 02 cm. Use energy considerations to find the aver- age frictional force that stops the bullet. Correct answer: 1713 . 51 N. Explanation: We can use conservation of energy to relate the initial kinetic energy of the bullet to the work done by the frictional force. 1 2 mv 2 = f s Solving for the frictional force, f , f = mv 2 2 s = (0 . 00283 kg)(270 m / s) 2 2(0 . 0602 m) = 1713 . 51 N . 002 (part 2 of 2) 10 points Assuming that the frictional force is constant, determine how much time elapsed between the moment the bullet entered the tree and the moment it stopped. Correct answer: 0 . 000445926 s. Explanation: If we assume the frictional force is constant, by Newtons second law we can assume con- stant deceleration, which implies v avg = v initial 2 . The time of deceleration is then t = s v avg = . 0602 m 270 m / s 2 = 0 . 000445926 s . keywords: 003 (part 1 of 1) 10 points A car of weight 2110 N operating at power 121 kW develops a maximum speed of 29 . 3 m / s on a level, horizontal road. Assume: The resistive force remains con- stant under the conditions mentioned below. What is the maximum speed of the car up an incline of 5 . 51 relative to the horizontal? Correct answer: 27 . 9298 m / s. Explanation: Basic Concepts: P = dW dt = ~ F ~v . Maximum Speed and Terminal Velocity Resistive forces. Let : W = 2110 N , p = 121 kW , v 1 = 29 . 3 m / s , and = 5 . 51 . Solution: On level ground, the only forces acting on the car in the horizontal direction are the applied force F 1 and the resistive force R . (Gravity and the normal force act perpen- dicular to this and add vectorially to zero.) Since the car has reached it maximum speed, it is no longer accelerating, so the net force is zero. That is F 1- R = 0 so R = F 1 . The applied force is related to the power through the relation P = ~ F 1 ~v 1 . The ap- plied force, ~ F 1 , is in the same direction as the velocity, so ~ F 1 ~v = F 1 v 1 . This means that F 1 = P v 1 = 121 kW 29 . 3 m / s 10 3 W kW = 4129 . 69 N . Nguyen, Thanh Homework 10 Due: Oct 9 2007, 7:00 pm Inst: D Weathers 2 When the car is on the incline, we now have...
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Homework_10 - Nguyen, Thanh Homework 10 Due: Oct 9 2007,...

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